Assume that the function is $f:\mathbb{R}²\to\mathbb{R}³$
Exercise: Find and classify the crititcal points of the function \begin{equation*} f(x,y)=\frac{xy}{2+x^4+y^4} \end{equation*}
Attempt Find $f'(x,y)$ og $f''(x,y)$:
\begin{equation*} \frac{\partial f}{\partial x}=\frac{y(2+x^4+y^4)-4x^4y}{(2+x^4+y^4)^2} \end{equation*} \begin{equation*} \frac{\partial f}{\partial y}=\frac{x(2+x^4+y^4)-4y^4x}{(2+x^4+y^4)^2} \end{equation*}
\begin{equation*} \frac{\partial^2 f}{\partial x^2}=-\frac{20yx^3}{(x^4+y^4+2)^2}+\frac{32x^7y}{(x^4+y^4+2)^3} \end{equation*}
\begin{equation*} \frac{\partial^2 f}{\partial y^2}=-\frac{20y^3x}{(x^4+y^4+2)^2}+\frac{32xy^7}{(x^4+y^4+2)^3} \end{equation*}
For $0=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$ we find
\begin{equation*} \frac{y(2+x^4+y^4)-4x^4y}{(2+x^4+y^4)^2}=\frac{x(2+x^4+y^4)-4y^4x}{(2+x^4+y^4)^2} \end{equation*}
\begin{equation*} y\left((2+x^4+y^4)-4x^4\right)=x\left((2+x^4+y^4)-4y^4\right) \end{equation*} A solution is therefore $x=y=0$
We would like to search for any other solutions within the parathesis
\begin{equation*} \begin{cases} 0=2+x^4+y^4-4x^4y\\ 0=2+x^4+y^4-4y^4x \end{cases} \end{equation*}
Gives: \begin{equation*} \begin{cases} 0=-4x^4y\\ 0=-4y^4x \end{cases} \end{equation*}
Problem: I know that there is for sure complex solutions to the system of equations. Although, since we are in the reals I have problems finding out weither:
- Since we are in the reals, we forget about the FTA, and then just take the solutions we can find without complex numbers
- There is something that I've simply missed? Although i have checked with WolframAlpha and used my common sense.
Thanks in advance
You have gone astray in several ways here. What you are looking for is points where both $\frac {\partial f}{\partial x} = 0$ and $\frac {\partial f}{\partial y} = 0$, but you've gotten distracted by where $\frac {\partial f}{\partial x} = \frac {\partial f}{\partial y}$, which indeed is necessary, but is not sufficient.
The set of equations you need to solve is $$\frac{y(2+x^4+y^4)-4x^4y}{(2+x^4+y^4)^2} = 0\\\frac{x(2+x^4+y^4)-4y^4x}{(2+x^4+y^4)^2} = 0$$ Multiplying through by the denominators, this is $$y(2+x^4+y^4)-4x^4y = 0\\x(2+x^4+y^4)-4y^4x = 0$$ If we let $x = 0$, then the system easily solves to $y = 0$, and vice versa. Since you've already found the $(0,0)$ solution, we can assume that both $x \ne 0, y\ne 0$ and so can divide them out, leaving
$$2-3x^4+y^4 = 0\\2+x^4-3y^4 = 0$$ This system is easy to solve to get $x^4 = 1, y^4 = 1$, which gives $$x = \pm 1, y = \pm 1$$
So there are 5 critical points in total.